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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
18% (02:55) correct
82%
(02:29)
wrong
based on 28
sessions
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Mapleman and Syrupo are two of the producers of maple syrup, each of which is either Canadian or American. Each producer competes with every other producer but with no others. 2/3 of Mapleman’s competitors are American, whereas more than 7/20 of Syrupo’s competitors are Canadian. What is the greatest possible number of maple syrup producers?
A. 58
B. 61
C. 64
D. 67
E. 68
Improved version !
A. 58
B. 61
C. 64
D. 67
E. 68
Improved version !
02 Apr 2026, 01:44
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This one's a nice counting problem. The concept being tested is setting up integer constraints from ratio conditions - classic GMAT Focus min/max territory.
Let N = total number of maple syrup producers.
Each producer competes with every other, so Mapleman has N-1 competitors and Syrupo has N-1 competitors.
Constraint from Mapleman: 2/3 of Mapleman's competitors are American. So N-1 must be divisible by 3. This means N ≡ 1 (mod 3).
Now there are two cases - Mapleman is either Canadian or American.
Case 1: Mapleman is Canadian.
Let A = number of American producers. All A of them are competitors of Mapleman, and 2/3 of Mapleman's (N-1) competitors are American, so A = 2(N-1)/3. Then C (total Canadians including Mapleman) = N - 2(N-1)/3 = (N+2)/3.
Case 2: Mapleman is American.
Then A-1 Americans compete with Mapleman, so A-1 = 2(N-1)/3, giving C = N - A = (N-1)/3. Now apply Syrupo's constraint: more than 7/20 of Syrupo's competitors are Canadian. For any sub-case of Case 2, C/N is small compared to A/N, so the fraction of Syrupo's Canadian competitors ends up being less than 7/20. Case 2 yields no solutions (you can verify it leads to 20 > 21, impossible).
Back to Case 1. Syrupo's competitors include (C-1) or C Canadians depending on whether Syrupo is Canadian or American. To maximize N, put Syrupo as American so all C Canadians compete with Syrupo:
C > 7(N-1)/20
(N+2)/3 > 7(N-1)/20
20(N+2) > 21(N-1)
20N + 40 > 21N - 21
61 > N
So N < 61. Combined with N ≡ 1 (mod 3), the greatest valid N is 58.
Check N = 58: A = 2(57)/3 = 38, C = 20. Syrupo's condition: 20 > 7(57)/20 = 19.95. Yes, 20 > 19.95. Valid.
N = 61 fails: C = 21, need 21 > 7(60)/20 = 21. But we need strictly greater than, so this fails.
Answer: A (58)
The trap is Case 2 - a lot of people try both cases but don't fully close out the algebra on Case 2 and end up second-guessing. Kill it fast and move on.
Let N = total number of maple syrup producers.
Each producer competes with every other, so Mapleman has N-1 competitors and Syrupo has N-1 competitors.
Constraint from Mapleman: 2/3 of Mapleman's competitors are American. So N-1 must be divisible by 3. This means N ≡ 1 (mod 3).
Now there are two cases - Mapleman is either Canadian or American.
Case 1: Mapleman is Canadian.
Let A = number of American producers. All A of them are competitors of Mapleman, and 2/3 of Mapleman's (N-1) competitors are American, so A = 2(N-1)/3. Then C (total Canadians including Mapleman) = N - 2(N-1)/3 = (N+2)/3.
Case 2: Mapleman is American.
Then A-1 Americans compete with Mapleman, so A-1 = 2(N-1)/3, giving C = N - A = (N-1)/3. Now apply Syrupo's constraint: more than 7/20 of Syrupo's competitors are Canadian. For any sub-case of Case 2, C/N is small compared to A/N, so the fraction of Syrupo's Canadian competitors ends up being less than 7/20. Case 2 yields no solutions (you can verify it leads to 20 > 21, impossible).
Back to Case 1. Syrupo's competitors include (C-1) or C Canadians depending on whether Syrupo is Canadian or American. To maximize N, put Syrupo as American so all C Canadians compete with Syrupo:
C > 7(N-1)/20
(N+2)/3 > 7(N-1)/20
20(N+2) > 21(N-1)
20N + 40 > 21N - 21
61 > N
So N < 61. Combined with N ≡ 1 (mod 3), the greatest valid N is 58.
Check N = 58: A = 2(57)/3 = 38, C = 20. Syrupo's condition: 20 > 7(57)/20 = 19.95. Yes, 20 > 19.95. Valid.
N = 61 fails: C = 21, need 21 > 7(60)/20 = 21. But we need strictly greater than, so this fails.
Answer: A (58)
The trap is Case 2 - a lot of people try both cases but don't fully close out the algebra on Case 2 and end up second-guessing. Kill it fast and move on.
